If I start with an $SU(5)$ gauge group and discover that the vacuum is preserved only by matrices of the form $G$ $$\begin{bmatrix} A & 0 \\ 0 & B \\ \end{bmatrix}$$ where the conditions on $A$ and $B$ ($A$ is a $3\times 3$ matrix and $B$ a $2\times 2$) are: that they must be Hermitian, and the matrix $G$ is traceless.
I don't understand why, if $tr(A)=−tr(B)$, then I have $U(1)$ and if I take $trA = trB = 0$ it is $SU(3)\times SU(2)$, so it means $SU(3)\times SU(2)\times U(1)$.
There is no "if $\operatorname {tr} G= 0= \operatorname {tr} A +\operatorname {tr} B$": you already assumed it.
The traceless matrix diag$(2I_3, -3I_2)$ manifestly commutes with all 12 linearly independent Gs, so it is a sole generator commuting with the full algebra; hence it generates a U(1) by exponentiation, relating to all remaining generators in G by a Cartesian $\times$.
The remaining 11 Gs may now be made traceless by subtracting a suitable multiple of this U(1) generator from G, $G'=G-(\operatorname {tr}A)\operatorname {diag}(I_3/3,-I_2/2)$. Now both the A's and the B's are traceless.
3×3 hermitean traceless matrices comprise the algebra of SU(3), and 2×2 such the algebra of SU(2), commuting between these blocks by construction/posit, above. You are then left with the direct sum of 1+8+3 generators of $SU(3)\times SU(2)\times U(1)$.
